Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 779.

## Sunday, July 8, 2012

### Problem 779: Triangle, Altitude, Orthocenter, Vertex, Midpoint, Side, Angle, 90 Degrees

Labels:
90,
altitude,
angle,
midpoint,
orthocenter,
perpendicular,
side,
triangle,
vertex

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By properties of orthocenter, and cyclic quadrilateral,

ReplyDeleteAngle DAC = Angle DHB

Thus,

Triangle DAC ~ Triangle DHB

Now, consider a rotation 90 degrees clockwise about point D,

DB maps to a line segment on DC,

DH maps to a line segment on DA.

So, after the above rotation,

triangle DHB will become the same position as triangle DAC, only with different size.

Hence, DM will map to a line segment on DF.

Which means, angle FDM = 90 degrees.

Triangle ADC is similar to triangle DHB

ReplyDeleteAnd since F is the mid point of AC and M is the mid point of BH, so triangle ADF is similar to DHM and DFC is similar to DHM

So angle FDM is same as angle ADC , which is 90

QED

http://img37.imageshack.us/img37/6047/problem779.png

ReplyDeleteDraw nine-points circle per attached sketch

This circle will pass through feet of altitudes, midpoint of each side and midpoint from orthocenter to each vertex.

Since ∠ (MNF)=90 => MF is a diameter

So ∠ (MDF)=90

∆ADC is rt angled at D, and F is the midpoint of the hypotenuse AC.

ReplyDeleteSo AF = DF, ∠ADF = A.

∆BDH is rt angled at D, and M is the midpoint of the hypotenuse BH.

So ∠BDM = ∠DBM = 90° - A (note BH ⊥ AC)

Follows ∠ADF + ∠BDM = 90°

Hence ∠FDM = 90°